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Leokris [45]
3 years ago
9

A solid metal sphere of radius a = 2.5 cm has a net charge Qin = - 3 nC (1 nC = 10-9C). The sphere is surrounded by a concentric

conducting spherical shell of inner radius b = 6 cm and outer radius c = 9 cm. The shell has a net charge Qout = + 2 nC. What is V0, the electric potential at the center of the metal sphere, given the potential at infinity is zero? V0 =
Physics
1 answer:
Furkat [3]3 years ago
4 0

Answer: 630 V

Explanation: In order to solve this problem we have to consider the potential given by:

In the region 0<a<b

V(r)= -∫E*dr where E can be calculated by Gaussian law then E= k*qa/r^2

where qa=-3 nC

then

The V(r)=k*qa/r+C C is zero since V(r=∞)=0

Finally we have

V(r)= k*qa (1/r)-(1/b)

thus V(a)= k*qa (1/a)-(1/b)

Replacing the values V(a) = 630 V, as the solid metal sphere is a equipotential object thus at the center have the same value of V that in r=a ( 630 V).

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It was once recorded that a Jaguar
Artyom0805 [142]

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

6 0
3 years ago
Why does the frequency of a siren get higher as an ambulance using that siren gets closer?
Archy [21]
The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.

7 0
3 years ago
Read 2 more answers
In an experiment to determine the s.h.c. of lead, a 0.80 kg block of lead is heated using a
gladu [14]

Answer:2.47

Explanation: did the math

8 0
2 years ago
Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c
Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0
3 years ago
12. The diameter of a circle is 2.42m. Calculate its<br>area in proper significant figure​
Sever21 [200]

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

A=\frac{\pi }{4} *D^{2}

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]

7 0
2 years ago
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